OFFSET
0,3
COMMENTS
Take two consecutive triangular numbers t1 and t2 and create a triangle using (0,0), (t1,t2) and (t2,t1). The area of this triangle will be ((n+1)^3)/2 for t1 = n*(n+1)/2. - J. M. Bergot, May 08 2012
LINKS
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-6,0,4,0,-1).
FORMULA
G.f.: (1+x+23x^2+22x^4+23x^5+x^7+x^8)/(1-x^2)^4.
a(n) = 0^n + n^3*(3/4 - (-1)^n/4).
a(n+1) = A129196(n)*(5/3 + (4/3)*cos(2*Pi*(n+1)/3)).
a(2n) = 4n^3, a(2n+1) = (2n+1)^3.
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 1 + 9*zeta(3)/8.
Sum_{n>=0} (-1)^n/a(n) = 1 - 5*zeta(3)/8. (End)
From Peter Bala, Jan 21 2024: (Start)
For n >= 1, a(n) = n*A129194(n) = n*Sum_{d divides n} (-1)^(d+1)*J(2,n/d), where the Jordan totient function J_2(n) = A007434(n). Cf. A309337.
Dirichlet g.f. for sequence without the a(0) term: (1 - 4/2^s)*zeta(s-3). - Peter Bala, Jan 21 2024
MATHEMATICA
Join[{1}, Table[GCD[Denominator[2 / n^3]], {n, 100}]] (* Vincenzo Librandi, Jul 26 2018 *)
PROG
(PARI) a(n) = if(n < 1, n==0, lcm(2, n^3)/2) \\ Andrew Howroyd, Jul 25 2018
(Magma) [1] cat [Denominator(2/n^3): n in [1..40]]; // Vincenzo Librandi, Jul 26 2018
CROSSREFS
KEYWORD
nonn,easy,mult
AUTHOR
Paul Barry, Apr 03 2007
EXTENSIONS
More terms from Vincenzo Librandi, Jul 26 2018
STATUS
approved